Chapter 7.1, Problem 11E

a.

To determine

To find: The velocity of the particle after 3 seconds.

Answer to Problem 11E

The velocity of the particle after 3 seconds is $-6$ feet per seconds.

Explanation of Solution

Given information:

A projectile is fired upward with initial velocity 90 feet per seconds.

Calculation:

The height of the projectile at any time $t$ can be given as $h\left(t\right)=-\frac{1}{2}\cdot 32t^2+90t$ or $h\left(t\right)=-16t^2+90t$ .

So, the velocity of the particle at any time $t$ is $v\left(t\right)=\frac{d}{dt}\left(-16t^2+90t\right)$ or $v\left(t\right)=-32t+90$ .

Substitute $t=3$ in the formula $v\left(t\right)=-32t+90$ to find the velocity after 3 seconds.

  $\begin{array}{c}v\left(3\right)=-32\times 3+90\\ =-96+90\\ =-6\end{array}$

Conclusion:

The velocity of the particle after 3 seconds is $-6$ feet per seconds.

b.

To determine

To find: The time when the particle hits the ground.

Answer to Problem 11E

The particle hits the ground after $5.625$ seconds.

Explanation of Solution

Given information:

A projectile is fired upward with initial velocity 90 feet per seconds.

Calculation:

The height of the projectile at any time $t$ can be given as $h\left(t\right)=-\frac{1}{2}\cdot 32t^2+90t$ or $h\left(t\right)=-16t^2+90t$ .

When particle hits the ground, the height will be 0.

So, solve the equation $-16t^2+90t=0$ .

  $\begin{array}{c}-16t^2+90t=0\\ t\left(-16t+90\right)=0\\ t=0,5.625\end{array}$

So, the particle hits the ground after $t=5.625$ seconds.

Conclusion:

The particle hits the ground after $5.625$ seconds.

c.

To determine

To find: The net distance traveled when the particle hits the ground.

Answer to Problem 11E

The net distance traveled by the particle is 0 feet.

Explanation of Solution

Given information:

A projectile is fired upward with initial velocity 90 feet per seconds.

Calculation:

The particle was initial fired upward from ground level and the particle returns to the round after $5.625$ seconds. So, the net distance traveled is 0 feet.

Conclusion:

The net distance traveled by the particle is 0 feet.

d.

To determine

To find: The total distance traveled when the particle hits the ground.

Answer to Problem 11E

The total distance traveled by the particle is 253.12 feet.

Explanation of Solution

Given information:

A projectile is fired upward with initial velocity 90 feet per seconds.

Calculation:

The height of the projectile at any time $t$ can be given as $h\left(t\right)=-\frac{1}{2}\cdot 32t^2+90t$ or $h\left(t\right)=-16t^2+90t$ .

So, the velocity of the particle at any time $t$ is $v\left(t\right)=\frac{d}{dt}\left(-16t^2+90t\right)$ or $v\left(t\right)=-32t+90$ .

The velocity is positive from $t=0$ sec to $t=2.813$ sec and negative from $t=2.813$ sec to $t=5.625$ sec.

So, the total distance traveled by the particle from $t=0$ sec to $t=5.625$ sec is:

  $\begin{array}{c}\text{Distance traveled}={\displaystyle {\int }_0^{2.813}\left(-32t+90\right)dt}-{\displaystyle {\int }_{2.813}^{5.625}\left(-32t+90\right)dt}\\ ={\left[-16t^2+90t\right]}_0^{2.813}-{\left[-16t^2+90t\right]}_{2.813}^{5.625}\\ =\left[\left(-16{\left(2.813\right)}^2+90\left(2.813\right)\right)-\left(-16{\left(0\right)}^2+90\left(0\right)\right)\right]\\ -\left[\left(-16{\left(5.625\right)}^2+90\left(2.813\right)\right)-\left(-16{\left(2.813\right)}^2+90\left(2.813\right)\right)\right]\\ =126.562496-\left(-126.562496\right)\\ =\text{253}\text{.124992}\\ =\text{253}\text{.12}\end{array}$

Conclusion:

The total distance traveled by the particle is 253.12 feet.